124. A 50.0-mL solution is initially 1.55% MgCl2 by mass and
has...

Question

124. A 50.0-mL solution is initially 1.55% MgCl2 by mass and 
has a density of 1.05 g>mL. What is the freezing point of the 
solution after you add an additional 1.35 g MgCl2? 
(Use i = 2.5 for MgCl2.)

Accepted Answer

Hi everyone here we have a question telling us that the density of a 100.0 mL solution of 2.55% calcium chloride by mass is 1.25 g per milliliter. If 1.25 g of calcium chloride was added to the solution, Calculate the new freezing point of the solution. The van Hoff factor for calcium chloride is 2.5. So first we need to calculate morality. Our morality equals the moles of the solute over kilograms of solvent. So our initial is 2.55%,, Which we are going to call 2.55 g In a 100 g solution. So the mass of our solution equals 100 0. millimeters Times 1. grams per mil leader Equals 102 .5 g of solution. So we have 100's who 0.5 g Times 2.55 grams of calcium chloride In 100 g of solution, Which equals 2. 1375 grams of calcium chloride. The mass of water equals the mass of our solution minus the mass of calcium chloride, which equals 102 0. g minus two point 61375 g of calcium chloride, Which equals 99 points eight 8625 g. Our mass of calcium chloride Equals 2. g Plus 1. g equals three .86375g. Our molar mass of calcium chloride equals 40 . grams per mole plus 35 0.45 grams. Permal times two Equals 110 . graham's Permal. So our morality Equal 0.03 four 82 moles of calcium chloride, Divided by 0.0 89 kg, Which equals 0. morality. Our change in temperature equals our temperature of our solvent minus the temperature of our solution. Our change in temperature equals R. Van Hoff factor times our freezing point constant times are morality. Our freezing point constant Equals 1. degree Celsius. Over morality. Our freezing point of our solvent Equals 0°C. So our change in freezing point Equals 2. Times 1.86 degree Celsius, divide by morality times 0.34 A six Morality Equals 1.62 Degrees C, which we're going to round to 1.6°C. So our change in freezing point Equals 0 -1.6°C Equals -1.6°C. And that is our final answer. Thank you for watching. Bye

124. A 50.0-mL solution is initially 1.55% MgCl2 by mass and has a density of 1.05 g>mL. What is the freezing point of the solution after you add an additional 1.35 g MgCl2? (Use i = 2.5 for MgCl2.)

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